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Theorem List for Metamath Proof Explorer - 11101-11200   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theorem4pos 11101 The number 4 is positive. (Contributed by NM, 27-May-1999.)
0 < 4

Theorem4ne0 11102 The number 4 is nonzero. (Contributed by David A. Wheeler, 5-Dec-2018.)
4 ≠ 0

Theorem5pos 11103 The number 5 is positive. (Contributed by NM, 27-May-1999.)
0 < 5

Theorem6pos 11104 The number 6 is positive. (Contributed by NM, 27-May-1999.)
0 < 6

Theorem7pos 11105 The number 7 is positive. (Contributed by NM, 27-May-1999.)
0 < 7

Theorem8pos 11106 The number 8 is positive. (Contributed by NM, 27-May-1999.)
0 < 8

Theorem9pos 11107 The number 9 is positive. (Contributed by NM, 27-May-1999.)
0 < 9

Theorem10posOLD 11108 The number 10 is positive. (Contributed by NM, 5-Feb-2007.) Obsolete version of 10pos 11500 as of 8-Sep-2021. (New usage is discouraged.) (Proof modification is discouraged.)
0 < 10

5.4.4  Some properties of specific numbers

This section includes specific theorems about one-digit natural numbers (membership, addition, subtraction, multiplication, division, ordering).

Theoremneg1cn 11109 -1 is a complex number (common case). (Contributed by David A. Wheeler, 7-Jul-2016.)
-1 ∈ ℂ

Theoremneg1rr 11110 -1 is a real number (common case). (Contributed by David A. Wheeler, 5-Dec-2018.)
-1 ∈ ℝ

Theoremneg1ne0 11111 -1 is nonzero (common case). (Contributed by David A. Wheeler, 8-Dec-2018.)
-1 ≠ 0

Theoremneg1lt0 11112 -1 is less than 0 (common case). (Contributed by David A. Wheeler, 8-Dec-2018.)
-1 < 0

Theoremnegneg1e1 11113 --1 is 1 (common case). (Contributed by David A. Wheeler, 8-Dec-2018.)
--1 = 1

Theorem1pneg1e0 11114 1 + -1 is 0 (common case). (Contributed by David A. Wheeler, 8-Dec-2018.)
(1 + -1) = 0

Theorem0m0e0 11115 0 minus 0 equals 0 (common case). (Contributed by David A. Wheeler, 8-Dec-2018.)
(0 − 0) = 0

Theorem1m0e1 11116 1 - 0 = 1 (common case). (Contributed by David A. Wheeler, 8-Dec-2018.)
(1 − 0) = 1

Theorem0p1e1 11117 0 + 1 = 1. (Contributed by David A. Wheeler, 7-Jul-2016.)
(0 + 1) = 1

Theorem1p0e1 11118 1 + 0 = 1. (Contributed by David A. Wheeler, 8-Dec-2018.)
(1 + 0) = 1

Theorem1p1e2 11119 1 + 1 = 2. (Contributed by NM, 1-Apr-2008.)
(1 + 1) = 2

Theorem2m1e1 11120 2 - 1 = 1. The result is on the right-hand-side to be consistent with similar proofs like 4p4e8 11149. (Contributed by David A. Wheeler, 4-Jan-2017.)
(2 − 1) = 1

Theorem1e2m1 11121 1 = 2 - 1 (common case). (Contributed by David A. Wheeler, 8-Dec-2018.)
1 = (2 − 1)

Theorem3m1e2 11122 3 - 1 = 2. (Contributed by FL, 17-Oct-2010.) (Revised by NM, 10-Dec-2017.) (Proof shortened by AV, 6-Sep-2021.)
(3 − 1) = 2

Theorem4m1e3 11123 4 - 1 = 3. (Contributed by AV, 8-Feb-2021.) (Proof shortened by AV, 6-Sep-2021.)
(4 − 1) = 3

Theorem5m1e4 11124 5 - 1 = 4. (Contributed by AV, 6-Sep-2021.)
(5 − 1) = 4

Theorem6m1e5 11125 6 - 1 = 5. (Contributed by AV, 6-Sep-2021.)
(6 − 1) = 5

Theorem7m1e6 11126 7 - 1 = 6. (Contributed by AV, 6-Sep-2021.)
(7 − 1) = 6

Theorem8m1e7 11127 8 - 1 = 7. (Contributed by AV, 6-Sep-2021.)
(8 − 1) = 7

Theorem9m1e8 11128 9 - 1 = 8. (Contributed by AV, 6-Sep-2021.)
(9 − 1) = 8

Theorem2p2e4 11129 Two plus two equals four. For more information, see "2+2=4 Trivia" on the Metamath Proof Explorer Home Page: mmset.html#trivia. This proof is simple, but it depends on many other proof steps because 2 and 4 are complex numbers and thus it depends on our construction of complex numbers. The proof o2p2e4 7606 is similar but proves 2 + 2 = 4 using ordinal natural numbers (finite integers starting at 0), so that proof depends on fewer intermediate steps. (Contributed by NM, 27-May-1999.)
(2 + 2) = 4

Theorem2times 11130 Two times a number. (Contributed by NM, 10-Oct-2004.) (Revised by Mario Carneiro, 27-May-2016.) (Proof shortened by AV, 26-Feb-2020.)
(𝐴 ∈ ℂ → (2 · 𝐴) = (𝐴 + 𝐴))

Theoremtimes2 11131 A number times 2. (Contributed by NM, 16-Oct-2007.)
(𝐴 ∈ ℂ → (𝐴 · 2) = (𝐴 + 𝐴))

Theorem2timesi 11132 Two times a number. (Contributed by NM, 1-Aug-1999.)
𝐴 ∈ ℂ       (2 · 𝐴) = (𝐴 + 𝐴)

Theoremtimes2i 11133 A number times 2. (Contributed by NM, 11-May-2004.)
𝐴 ∈ ℂ       (𝐴 · 2) = (𝐴 + 𝐴)

Theorem2txmxeqx 11134 Two times a complex number minus the number itself results in the number itself. (Contributed by Alexander van der Vekens, 8-Jun-2018.)
(𝑋 ∈ ℂ → ((2 · 𝑋) − 𝑋) = 𝑋)

Theorem2div2e1 11135 2 divided by 2 is 1 (common case). (Contributed by David A. Wheeler, 8-Dec-2018.)
(2 / 2) = 1

Theorem2p1e3 11136 2 + 1 = 3. (Contributed by Mario Carneiro, 18-Apr-2015.)
(2 + 1) = 3

Theorem1p2e3 11137 1 + 2 = 3 (common case). (Contributed by David A. Wheeler, 8-Dec-2018.)
(1 + 2) = 3

Theorem3p1e4 11138 3 + 1 = 4. (Contributed by Mario Carneiro, 18-Apr-2015.)
(3 + 1) = 4

Theorem4p1e5 11139 4 + 1 = 5. (Contributed by Mario Carneiro, 18-Apr-2015.)
(4 + 1) = 5

Theorem5p1e6 11140 5 + 1 = 6. (Contributed by Mario Carneiro, 18-Apr-2015.)
(5 + 1) = 6

Theorem6p1e7 11141 6 + 1 = 7. (Contributed by Mario Carneiro, 18-Apr-2015.)
(6 + 1) = 7

Theorem7p1e8 11142 7 + 1 = 8. (Contributed by Mario Carneiro, 18-Apr-2015.)
(7 + 1) = 8

Theorem8p1e9 11143 8 + 1 = 9. (Contributed by Mario Carneiro, 18-Apr-2015.)
(8 + 1) = 9

Theorem9p1e10OLD 11144 9 + 1 = 10. (Contributed by Mario Carneiro, 18-Apr-2015.) Obsolete version of 9p1e10 11481 as of 8-Sep-2021. (New usage is discouraged.) (Proof modification is discouraged.)
(9 + 1) = 10

Theorem3p2e5 11145 3 + 2 = 5. (Contributed by NM, 11-May-2004.)
(3 + 2) = 5

Theorem3p3e6 11146 3 + 3 = 6. (Contributed by NM, 11-May-2004.)
(3 + 3) = 6

Theorem4p2e6 11147 4 + 2 = 6. (Contributed by NM, 11-May-2004.)
(4 + 2) = 6

Theorem4p3e7 11148 4 + 3 = 7. (Contributed by NM, 11-May-2004.)
(4 + 3) = 7

Theorem4p4e8 11149 4 + 4 = 8. (Contributed by NM, 11-May-2004.)
(4 + 4) = 8

Theorem5p2e7 11150 5 + 2 = 7. (Contributed by NM, 11-May-2004.)
(5 + 2) = 7

Theorem5p3e8 11151 5 + 3 = 8. (Contributed by NM, 11-May-2004.)
(5 + 3) = 8

Theorem5p4e9 11152 5 + 4 = 9. (Contributed by NM, 11-May-2004.)
(5 + 4) = 9

Theorem5p5e10OLD 11153 5 + 5 = 10. (Contributed by NM, 5-Feb-2007.) Obsolete version of 5p5e10 11581 as of 8-Sep-2021. (New usage is discouraged.) (Proof modification is discouraged.)
(5 + 5) = 10

Theorem6p2e8 11154 6 + 2 = 8. (Contributed by NM, 11-May-2004.)
(6 + 2) = 8

Theorem6p3e9 11155 6 + 3 = 9. (Contributed by NM, 11-May-2004.)
(6 + 3) = 9

Theorem6p4e10OLD 11156 6 + 4 = 10. (Contributed by NM, 5-Feb-2007.) Obsolete version of 6p4e10 11583 as of 8-Sep-2021. (New usage is discouraged.) (Proof modification is discouraged.)
(6 + 4) = 10

Theorem7p2e9 11157 7 + 2 = 9. (Contributed by NM, 11-May-2004.)
(7 + 2) = 9

Theorem7p3e10OLD 11158 7 + 3 = 10. (Contributed by NM, 5-Feb-2007.) Obsolete version of 7p3e10 11588 as of 8-Sep-2021. (New usage is discouraged.) (Proof modification is discouraged.)
(7 + 3) = 10

Theorem8p2e10OLD 11159 8 + 2 = 10. (Contributed by NM, 5-Feb-2007.) Obsolete version of 8p2e10 11595 as of 8-Sep-2021. (New usage is discouraged.) (Proof modification is discouraged.)
(8 + 2) = 10

Theorem1t1e1 11160 1 times 1 equals 1. (Contributed by David A. Wheeler, 7-Jul-2016.)
(1 · 1) = 1

Theorem2t1e2 11161 2 times 1 equals 2. (Contributed by David A. Wheeler, 6-Dec-2018.)
(2 · 1) = 2

Theorem2t2e4 11162 2 times 2 equals 4. (Contributed by NM, 1-Aug-1999.)
(2 · 2) = 4

Theorem3t1e3 11163 3 times 1 equals 3. (Contributed by David A. Wheeler, 8-Dec-2018.)
(3 · 1) = 3

Theorem3t2e6 11164 3 times 2 equals 6. (Contributed by NM, 2-Aug-2004.)
(3 · 2) = 6

Theorem3t3e9 11165 3 times 3 equals 9. (Contributed by NM, 11-May-2004.)
(3 · 3) = 9

Theorem4t2e8 11166 4 times 2 equals 8. (Contributed by NM, 2-Aug-2004.)
(4 · 2) = 8

Theorem5t2e10OLD 11167 5 times 2 equals 10. (Contributed by NM, 5-Feb-2007.) Obsolete version of 5t2e10 11619 as of 8-Sep-2021. (New usage is discouraged.) (Proof modification is discouraged.)
(5 · 2) = 10

Theorem2t0e0 11168 2 times 0 equals 0. (Contributed by David A. Wheeler, 8-Dec-2018.)
(2 · 0) = 0

Theorem4d2e2 11169 One half of four is two. (Contributed by NM, 3-Sep-1999.)
(4 / 2) = 2

Theorem2nn 11170 2 is a positive integer. (Contributed by NM, 20-Aug-2001.)
2 ∈ ℕ

Theorem3nn 11171 3 is a positive integer. (Contributed by NM, 8-Jan-2006.)
3 ∈ ℕ

Theorem4nn 11172 4 is a positive integer. (Contributed by NM, 8-Jan-2006.)
4 ∈ ℕ

Theorem5nn 11173 5 is a positive integer. (Contributed by Mario Carneiro, 15-Sep-2013.)
5 ∈ ℕ

Theorem6nn 11174 6 is a positive integer. (Contributed by Mario Carneiro, 15-Sep-2013.)
6 ∈ ℕ

Theorem7nn 11175 7 is a positive integer. (Contributed by Mario Carneiro, 15-Sep-2013.)
7 ∈ ℕ

Theorem8nn 11176 8 is a positive integer. (Contributed by Mario Carneiro, 15-Sep-2013.)
8 ∈ ℕ

Theorem9nn 11177 9 is a positive integer. (Contributed by NM, 21-Oct-2012.)
9 ∈ ℕ

Theorem10nnOLD 11178 Obsolete version of 10nn 11499 as of 6-Sep-2021. (Contributed by NM, 8-Nov-2012.) (New usage is discouraged.) (Proof modification is discouraged.)
10 ∈ ℕ

Theorem1lt2 11179 1 is less than 2. (Contributed by NM, 24-Feb-2005.)
1 < 2

Theorem2lt3 11180 2 is less than 3. (Contributed by NM, 26-Sep-2010.)
2 < 3

Theorem1lt3 11181 1 is less than 3. (Contributed by NM, 26-Sep-2010.)
1 < 3

Theorem3lt4 11182 3 is less than 4. (Contributed by Mario Carneiro, 15-Sep-2013.)
3 < 4

Theorem2lt4 11183 2 is less than 4. (Contributed by Mario Carneiro, 15-Sep-2013.)
2 < 4

Theorem1lt4 11184 1 is less than 4. (Contributed by Mario Carneiro, 15-Sep-2013.)
1 < 4

Theorem4lt5 11185 4 is less than 5. (Contributed by Mario Carneiro, 15-Sep-2013.)
4 < 5

Theorem3lt5 11186 3 is less than 5. (Contributed by Mario Carneiro, 15-Sep-2013.)
3 < 5

Theorem2lt5 11187 2 is less than 5. (Contributed by Mario Carneiro, 15-Sep-2013.)
2 < 5

Theorem1lt5 11188 1 is less than 5. (Contributed by Mario Carneiro, 15-Sep-2013.)
1 < 5

Theorem5lt6 11189 5 is less than 6. (Contributed by Mario Carneiro, 15-Sep-2013.)
5 < 6

Theorem4lt6 11190 4 is less than 6. (Contributed by Mario Carneiro, 15-Sep-2013.)
4 < 6

Theorem3lt6 11191 3 is less than 6. (Contributed by Mario Carneiro, 15-Sep-2013.)
3 < 6

Theorem2lt6 11192 2 is less than 6. (Contributed by Mario Carneiro, 15-Sep-2013.)
2 < 6

Theorem1lt6 11193 1 is less than 6. (Contributed by NM, 19-Oct-2012.)
1 < 6

Theorem6lt7 11194 6 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
6 < 7

Theorem5lt7 11195 5 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
5 < 7

Theorem4lt7 11196 4 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
4 < 7

Theorem3lt7 11197 3 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
3 < 7

Theorem2lt7 11198 2 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
2 < 7

Theorem1lt7 11199 1 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
1 < 7

Theorem7lt8 11200 7 is less than 8. (Contributed by Mario Carneiro, 15-Sep-2013.)
7 < 8

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